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MICROPROCESSOR & INTERFACING LAB

(EE-329-F)

LABORATORY MANUAL

V - SEMESTER

Prepared by

VIKRANT VERMA(A.P.)

BRCM College of Engineering and Technology

Department of Electrical & Electronics &Engineering

Bahal, Bhiwani - 127028

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LIST OF EXEPRIMENTS

S.No Name of Experiments Page No

1. Introduction of microprocessor 8085 trainer kit – 85AD

2. Write an ALP to perform the addition of two 8 bit numbers.

3. Write an ALP to perform the subtraction of two 8 bit numbers.

4. Write an ALP to perform the addition with carry of two 8 bit numbers.

5. Write an ALP to perform the subtraction with barrow of two 8 bit numbers.

6. Write an ALP to perform the addition of two BCD numbers.

7. Write an ALP to perform the subtraction of two BCD numbers.

8. Write an ALP to perform the multiplication of two 8 bit numbers by repeated

addition method.

Write an ALP to perform the multiplication of two 8 bit numbers by bit

9. Rotation method.

Write an ALP to perform the division of two 8 bit numbers by repeated

10. addition method.

Write an ALP to perform the division of two 8 bit numbers by bit rotation

11. method.

12. Write an ALP to find the square of given numbers in array.

13. Write an ALP to find largest number in an array.

14. Study of 8086 microprocessor kit

15. Write an ALP to perform addition of two 16 bit numbers.

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EXPERIMENT NO.1

AIM:

To study about introduction of microprocessor 8085 trainer kit - 85AD. APPARATUS REQUIRED:

8085 - Microprocessor kit.

THEORY:

The system has got 8085 as the Central Processing Unit. The clock frequency for the system is

3.0MHz and is generated from a crystal of 6.14MHz. 8085 has got 8 bit data lines and 16 bit address

lines. The lower 8 address lines and 8 bit data lines are multiplexed. Since the lower 8 address bits

appear on the bus during the first clock cycle of a machine cycle and the 8 bit data appears on the

bus during the 2nd and 3rd clock cycle, it becomes necessary to latch the lower 8 address bits during

the first clock cycle so that the 16 bit address remains available in subsequent cycles. This is

achieved using a latch 74LS373.

The training kit which we are going to use in this lab is STUDENT-85AD which

communicates with the outside world through a general purpose IBM PC Compatible ASCII

keyboard and 16x2 Liquid Crystal Display (LCD). The kit also has the capability of interfacing with

CRT terminal through the interface provided on the board.

The on board resident system monitor software is very powerful and provides various

software utilities. The kit provides support for powerful software commands like INSERT,

DELETE, BLOCK MOVE, RELOCATE, STRING, FILL and MEMORY COMPARE etc. The kit

is configured around the internationally adopted standard STD bus which is most popular bus for

process control and real time applications. All the address, data and control lines are available at the

edge connector. The kit is fully expandable for any kind of application. MEMORY:

8085 kit provides 8/32K bytes of RAM using 6264/62256 chip and 8K bytes of EPROM for

monitor. There is one memory space provided on kit. This one space can be defined any address

slots from 8000 - DFFF depending upon the size of the memory chip to be used. Total onboard

memory can be extended to 64K bytes. I/O DEVICES

The various I/O chips used in STUDENT-85AD microprocessor kit are 8255, 8253 & 8155. The functional role of all these chips is given below:

8255(Programmable Peripheral Interface)

8255 is a programmable peripheral interface (PPI) designed to use with 8085 Microprocessor.

This basically acts as a general purpose I/O device to interface peripheral equipments to the system

bus. It is not necessary to have an external logic to interface with peripheral devices since the

functional configuration of 8255 is programmed by the system software. It has got three Input/Output

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ports of 8 lines each (PORT-A, PORT-B & PORT-C). Port C can be divided into two ports of 4 lines

each named as Port C upper and Port C lower. Any Input output combination of Port A, Port B, Port

C upper and lower) can be defined using the appropriate software commands. The kit provides 24

Input/output ports using 8255 chips. 8253(Programmable Internal Timer)

This chip is a programmable interval Timer/Counter and can be used for the generation of

accurate time delays under software control. Various other functions that can be implemented with

this chip are programmable rate generator, Even Counter, Binary rate Multiplier, Real Time Clock

etc. This chip has got three in dependent 16 bit counters each having a count rate of up to 2 KHz. The

first Timer/Counter (i.e. Counter 0) is being used for Single Step operation. However, its connection

are also brought at connector space C4. For single step operation CLKO signal of Counter 0 is getting

a clock frequency of 1.535 MHz. The counter 1 is. used to generate clock for 8251. Counter 1 &

Counter 2 are free for the user. Clock for the CLK1, CLK2 is to be given externally. 8155 (Programmable I/O Port & Timer Interface) –Optional

8155 is a programmable I/O ports and timer interface designed to use with 8085

Microprocessor. The 8155 includes 256 bytes of R/W memory, three I/O ports and a Timer. This

basically acts as a general purpose I/O device to interface peripheral equipments to the system bus. It

is not necessary to have an external logic to interface with peripheral devices since the functional

configuration of 8155 is programmed by the system software. It has got two 8-bit parallel I/O port

(Port-A, Port-B) and one 6-bit (Port-C). Ports A & B also can be programmed in the handshake mode,

each port using three signals as handshake signals from Port-C. The timer is a 14 bit down counter

and has four modes.

List of ASCII Keyboard Commands

Sr. Command Description Command Syntax

No.

1. M Examine/ Modify [M]<ADDRESS>[DATA] [DATA]….[,]

Memory

2. E Enter a memory [E]<ADDRESS>[DATA] [DATA]….[$]

block

3. R Examine/ Modify [R] <REG. IDENTIFIER> [$]

register

4. S Single Step [S] <Starting Address>[,]

5. G Go [G] < Starting Address>[$]

6. B Block Move [B]< Starting Address of source>[,]< End

Address of source >[,]< Starting Address of

destination>[$]

7. I Insert [I]< Starting Address of the program>[,]<

End Address of the program >[,]<Address

from where the byte or bytes are to be

entered>[,]<No. of bytes>[,][DATA][$]

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8. D Delete [D]<Starting Address of the

program>[,]<Ending address of the

program>[,]<starting address from where the

bytes are to be deleted>[,]<Ending address

till where the bytes are to be deleted>[$]

9. N Insert Data [N]< Starting Address of the program/ data

area >[,] <Ending Address of the

program>[,]< Starting Address at which the

bytes are to be entered>[,]<No. of

bytes>[,][DATA][.][$]

10. O Delete Data [O]< Starting Address of the program or data

area>[,]< End Address of the program/ data

area >[,]< Starting Address from where the

deletion should start>[,]<End address till

where bytes are to be deleted>[$]

11. F Fill [F]< Starting Address of program/ data

area>[,]<End Address>[,]<Constant to be

filled>[$]

12. H Relocate [H]< Starting Address of the

program>[,]<End Address of the

program>[,]<Destination Address>[$]

13. J Memory Compare [J]< Starting Address of the first

block>[,]<End Address of thefirst

block>[,]< Starting Address of second

block>[$]

14. K String [K]< Starting Address of the

program>[,]<End address of the

program>[,]<Address of the location at

which first byte of the string lies>[,]<Address

of the location at which last byte of the string

lies>[$]

Table: 1 Details of the commands used in STUDENT-85AD

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Sr. No. Register Identifier Register Name

1. A Register A or accumulator

2. B Register B

3. C Register C

4. D Register D

5. E Register E

6. F Register F

7. I Interrupt Mask Register

8. H Register H

9. L Register L

10. S Stack Pointer MSB

11. P Program Counter MSB

Table: 2 Details of the Register Identities used in STUDENT-85AD Modes of Operation:

In STUDENT-85AD there is an onboard facility of assembler/ disassemble. On pressing ‘1’

key, STUDENT-85AD comes in assembler/ disassemble mode depending upon the next key you

press i.e.:

‘A’- Assembler mode

‘C’- Disassembler mode

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Assembler Mode:

On pressing the key ‘A’ STUDENT-85AD comes into the assembler mode. As soon as ‘A’ is

pressed, kit asks RAM address. This will be the starting address of the program to be entered. After

entering the starting address, press <CR> key, it displays the entered starting address in the upper line

of the LCD screen. Now it waits for the mnemonics entry.

One can enter all the valid mnemonics of 8085 and the Pseudo commands. If the entered

alphabets do not form a valid mnemonics or a Pseudo command, the carriage goes to same line and

prints the address of the previous line. Hence the entry of the wrong mnemonic is indicated by giving

the same line to the user.

Entry of a space completes one field of entry: and processing of that field is done immediately by the command. By field, we mean, mnemonic as one field, operand or label as another field.

Using this mode one can write or feed the program using assembly language mnemonics. Format:

[1]<ENTER> [A] <ENTER> The display will show

RAM ADRR:

Type the desired RAM ADDRESS e.g. 2000, the display looks like figure below

RAM ADRR: 2000

(Note: RAM Address is always a16-bit hexadecimal number.)

Press <ENTER> and the display will show the RAM Address blank against it.

2000:

One can type the command mnemonics in blank space. After that on pressing ENTER or SPACE key, the RAM Address will be automatically incremented. In case of invalid mnemonics the monitor software will erase the mnemonic and will remain on the same RAM Address until the command typed is a valid mnemonic. Then the display will appear like figure below:

2000: MOV B, A

In this manner one can type all a whole program which automatically gets stored in the RAM of the kit and RUN the program using ‘GO’ command thereafter. DISASSEMBLER MODE:

Disassembler mode can be selected by pressing the ‘1’ and then ‘C’. This command disassembles the program as specified by the STARTING address and END address. In case one

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wants to proceed further, press <CR> key, otherwise <Esc> key will exit from the disassemble mode. MEMORY MAPPING:

STUDENT-85AD kit provides 8/32 KB of RAM and 16 KB of EPROM. The total onboard

memory can be expanded to 64 KB. For the system operation the monitor should start from address

0000H. A minimum of 8KB RAM should be there on the board with starting address as 2000H. INPUT/ OUTPUT MAPPING:

Device Active range Port Port Numbers Selected Device

Addresses

8255-I 00-07 PPI

00 and 04 Port-A 01 and 05 Port-B

02 and 06 Port-C 03 and 07 Control Word

8255-II 08-0B PPI 08 Port-A

09 Port-B 0A Port-C

0B Control Word 8253 10-17 PIT

10 and 14 Counter 0 11 and 15 Counter 1

12 and 16 Counter 2 13 and 17 Control Word

LCD 38-3F

Developing/ Debugging Software:

STUDENT-85AD kit provides software features like Relocate, String, Insert, Delete,

Assembler, Disassembler, Programming etc. which find extensive application in developing/

debugging software. The various steps involved in developing software are: 1. Define the problem in the form of a flow chart. 2. Write the program in Assembly Language of 8085. 3. Assemble the program through Assembler command. 4. Enter the program in RAM area and RUN it.

It is likely that the program may not run in one shot because some mistakes can be there in it. The process of finding these mistakes and removing them is called the debugging of the program.

One way of entering the program is in HEX code of the mnemonics and the other way is

through assembler. In assembler mode you can write the program in mnemonics form and inspect the

disassembly form with its HEX code.

One way of finding the mistakes in the program is to run the program in single instruction mode and after each step compare what the program is doing and what is it supposed to do. In the

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process of this one might have to examine the contents of the memory locations or the content of internal registers after the execution of each instruction.

During this process of debugging, at some time user might just like to examine the status of

the program at a particular point. If this point is near the beginning of the program, one can reach this

point by single instruction facility. But if the point is quite far from the beginning of the program, it is

time saving to make use of BREAK POINT facility. For this introduce a RST5 instruction (EF) at the

point to be examined and run the program at full speed using ‘GO’ command. When during the

execution of the program, this instruction is encountered; the control of the processor is transferred to

the monitor. The monitor saves the user registers and displays a sign ‘STUDENT-85AD’ on the LCD

screen. No one can examine the status of any memory location or any internal register. One can

change the content of memory location or register if necessary.

Sometimes while debugging user may find that certain instructions are to be added to the

program or to be deleted from the program. The program written for one memory area can be made

operative for some other area using the RELOCATE command.

Sometimes it is required to execute the program in single cycle mode. For this one can make use of single cycle facility (optional) on the board of STUDENT-85AD kit.

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EXPERIMENT NO. 2 Aim

Write a well-documented program using 8085 for addition of two 8-bit numbers. Apparatus

8085 microprocessor kit (STUDENT-85AD), ASCII (PS-2) keyboard. Theory :-

2501H is the address of memory location for the 1st number. 2nd

number is stored at next

memory location. By the virtue of the program given below we first move the 1st number into

accumulator register then add the 2nd

number to this and use the next memory location i.e. 2503h to

store the result. The program given below is self explanatory.

Program:-

MEMORY MACHINE MNEMONICS OPERANDS COMMENTS

2000 21,01,25 LXI H,2501 Get address of 1st

no. in HL pair

2003 7E MOV A,M Move 1st

no. in accumulator

2004 23 INX H HL points the address 2502H

2005 86 ADD M Add the 2nd no.

2006 23 INX H HL points 2503H

2007 77 MOV M,A Store result in 2503H.

2008 CF RST 1 Terminate

Procedure:-

(1) Connect the kit to the power supply. (2) Connect the PS-2 keyboard to the PS-2 (Female) connector CN-7 of the kit. (3) Switch ON the kit. (4) LCD displays “STUDENT-85”, if not so check power connections.

(5) If backlight is glowing and display shows something else press the RESET key provided on

the kit. If still the persists contact Lab staff.

(6) To Feed the program and data in the RAM, for machine language follow the steps 7 to 9 and for assembly language programming steps 10 to 14.

(7) Press [M]<Starting Address >[ENTER]. Starting address here is 2000H, so syntax will be

[M]<2000>[ENTER].

(8) The LCD display will show the address and its content. Modify the content according to the

program using keyboard then press [ENTER] or [SPACEBAR]. The content will be saved and

display will move the next address. Modify if required. (9) Use UP/DOWN arrow keys move between the memories locations. (10) Press [1][A] to select assembler mode.

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(11) The display will show ‘RAM ADR: [BLANK]’, asking for the starting address which is

2000H in the above program.

(12) Type [2000] and press [ENTER].

(13) Display will show ‘2000: [BLANK]’. Type the mnemonics of the command operand etc. and

press [ENTER].

(14) The display will show the next location. Type desired mnemonics and operand. Keep on doing this until the end of the program.

(15) Feed the data using the [M] command. (16) To RUN/ EXECUTE the program Press [G]<Starting Address>[$]. (17) Go to memory locations used for result storage using [M] command and verify the results.

Figure: Flow-Chart of the Process

START

Get the first No.

Get the second No.

Add. Two numbers

Store the result

END

Input Data: 2501- 67h

2502- 22h Output Data: 2503- 89h Precautions:-

(1) Make sure proper handling of equipments/ kits.

(2) Make sure that all the machine codes/ mnemonics are as per the program. Result:-

Addition of 67H and 22H is 89H. Hence experiment for the addition of two 8-bit numbers has

been successfully performed.

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EXPERIMENT NO. 3 Aim:-

Write a program using 8085 for subtraction of two 8-bit numbers. Apparatus:-

8085 microprocessor kit (STUDENT-85AD), ASCII (PS-2) keyboard. Theory:-

2501h is the address of memory location for the 1st number. 2nd



accumulator register then subtract the 2nd

number from this and use the next memory location

i.e. 2503h to store the result. The program given below is self explanatory. Program;-

Memory Opcode Mnemonics Operands Comments

address

2000 21,01,25 LXI H, 2501 Get address of 1st no. in HL pair

2003 7E MOV A, M Move 1st no. in accumulator

2004 23 INX H HL points 2502H.

2005 96 SUB M Subtract 2nd no. from 1st no.

2006 23 INX H HL points 2503H.

2007 77 MOV M, A Move contents of acc. to memory

2008 CF RST 1 Stop

Procedure:-




(6) To feed the program and data in the RAM, for machine language follow the steps 7 to 9 and for assembly language programming steps 10 to 14.


[M]<2000>[ENTER].



display will move the next address. Modify if required. (9) Use UP/DOWN arrow keys move between the memory locations. (10) Press [1][A] to select assembler mode.

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(11) The display will show ‘RAM ADR: [BLANK]’, asking for the starting address which is




press [ENTER].



Flow-Chart of the process :-

Start

Get 1st number from memory

Get 2nd number from memory

Subtract 2nd number from the 1st

Store result

Stop

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Input Data: Case-I 2501h- 67h, 2502h- 22h

Case-II 2501h- 22h, 2502h- 67h Output Data: Case-I 2503h- 45h

Case-II 2503h- BBh Precautions:-



Subtraction of 22h from 67h is 45h. If we are subtracting a bigger number from a smaller

number i.e. 22h-67h, then result is two’s complement of 45h i.e. BBh. Hence experiment for the

subtraction of two 8-bit numbers has been successfully performed.

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Write a well-documented program using 8085 for addition of two 8-bit numbers with carry. Apparatus:-



number is stored at

next memory location. By the virtue of the program given below we first move the 1st number

into accumulator register then add the 2nd

number to this and if a carry is there we use C

register to store carry. We use next two memory locations i.e. 2503h & 2504h to store the

result and carry respectively. The program given below is self explanatory. Program:-

Memory Machine Labels Mnemonics Operands Comments

Address Code

2000 21,01,25 LXI H,2501 Get address of 1st no. in HL pair

2003 MVI C,00 MSB of sum, Initial value=00


no. in accumulator


2007 86 ADD M Add the 2nd

no.

2008 D2,0D,25 JNC AHEAD If no carry jump to AHEAD

200B 0C INR C If carry increment C

200C 32,03,25 AHEAD STA 2503h Store result at 2503h.

200F 79 MOV A,C Get carry in acc.

2010 32,04,25 STA 2504h Store carry at 2504h

2013 76 HLT Stop

Procedure:-






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[M]<2000>[ENTER].



display will move the next address. Modify if required. (9) Use UP/DOWN arrow keys move between the memory locations. (10) Press [1][A] to select assembler mode. (11) The display will show ‘RAM ADR: [BLANK]’, asking for the starting address which is




press [ENTER].




Start



Add 2nd number to the 1st

No Is there a Carry?

Yes

Increment C reg.

Store result

Stop

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2502- 22h

Output Data: 2503- 89h Precautions:-



Addition of 67h and 22h is 89h. Hence experiment for the addition of two 8-bit numbers has


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EXPERIMENT NO. 5 Aim:

Write a well-documented program using 8085 for subtraction of two 8-bit numbers with carry. Apparatus:






number from this and if a borrow is there we use C

register to store it. We use next two memory locations i.e. 2503h & 2504h to store the result

and borrow respectively. The program given below is self explanatory. Program:-


Address Code

2000 21,01,25

LXI H,2501 Get address of 1st no. in HL

pair.

2003 0E,00 MVI C,00 MSB of sum, Initial value=00


no. in accumulator


2007 86 SUB M Subtract 2nd

no. from acc.

2008 D2,0D,25 JNC AHEAD If no borrow jump to AHEAD

200B 0C INR C If borrow increment C

200C 32,03,25 AHEAD STA 2503h Store result at 2503h.

200F 79 MOV A,C Get carry in acc.

2010 32,04,25 STA 2504h Store carry at 2504h

2013 76 HLT Stop

Procedure:-

(1) Connect the kit to the power supply. (2) Connect the PS-2 keyboard to the PS-2 (Female) connector CN-7 of the kit. (3) Switch ON the kit. (4) LCD displays “STUDENT-85”, if not so check power connections. (5) If backlight is glowing and display shows something else press the RESET key provided on

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[M]<2000>[ENTER].







press [ENTER].




2502- 22h

Output Data: 2503- 89h

Precautions:-


(2) Make sure that all the machine codes/ mnemonics are as per the program.

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Start



Subtract 2nd number from 1st

Is there a No borrow?

Yes

Increment C reg.

Store result

Stop Result:-

Addition of 67h and 22h is 89h. Hence experiment for the addition of two 8-bit numbers has


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EXPERIMENT NO. 6

Aim:-

Write a program using 8085 for addition of two BCD (8-bit) numbers.

Apparatus:-

8085 microprocessor kit (STUDENT-85AD), PS-2 keyboard.

Theory:




accumulator register then add the 2nd

number to this and use the DAA (Decimal Adjust Accumulator)

instruction to convert result to BCD. The result is then stored into the memory location 2503h. The

comments in program give explanation of what happens after the execution of that particular

instruction. Procedure:-






[M]<2000>[ENTER].







press [ENTER].



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Program;-

Memory Op-code Labels Mnemonics Operands Comments

address

2000 21,01,25

LXI H, 2501 Get address of 1st no. in H-L

Pair

2003 0E,00

MVI C,00h Initialize C reg. to store MSB of

result.

2005 7E MOV A, M Move 1st no. in Accumulator

2006 23 INX H H-L points to 2502h.

2007 86 ADD M Add 2nd

no. to 1st no.

2008 27 DAA Decimal Adjust Accumulator

2009 D2,0D,20 JNC AHEAD If no carry go to AHEAD

200C 0C INR C If carry increment C reg.

200D 32,03,25 AHEAD STA 2503h LSB of the sum in 2503h.

2010 79 MOV A,C MSB of sum in acc.

2011 32,04,25 STA 2504h Store MSB at 2504h

2014 76 HLT Stop

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Flow-Chart of the process :

Start


Initialize reg. C for Carry


Add 1st and 2nd number

Decimal Adjust Accumulator

Yes Is there a

carry ?

Increment C register

Store result Stop

Input Data: 2501- 67h, 2502- 22h

Output Data: 2503- 89h

Precautions:-



Addition of 22h from 67h is 89h. Hence experiment for the addition of two 8-bit BCD numbers has been successfully performed.

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EXPERIMENT NO. 7

Aim:-

Write a program using 8085 for subtraction of two BCD (8-bit) numbers. Apparatus:-

8085 microprocessor kit (STUDENT-85AD), PS-2 keyboard. Theory:-





number from this and use the DAA (Decimal Adjust

Accumulator) instruction to convert result to BCD. The result is then stored into the memory location

2503h. The comments in program give explanation of what happens after the execution of that

particular instruction. Program;-

Memory Op-code Mnemonics Operands Comments

address

2000 21,02,25 LXI H, 2502 Get address of 2nd

no. in H-L pair

2003 3E,99 MVI A,99h Place 99h in acc.

2005 96 SUB M 9’s complement of 2nd

number

2006 3C INR A 10’s complement of 2nd

number

2007 2B DCX H Decrement H-L pair

2008 86 ADD M Add 1st

number to 10’s complement of 2nd

number

2009 27 DAA Decimal Adjust Accumulator

200A 32,03,25 STA 2503h Store result at 2503h

200D CF RST 1 Stop

Procedure:-






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[M]<2000>[ENTER].







press [ENTER].




Start


2’s complement of the number


Add 1st no. + 2’s complement of

Decimal Adjust Accumulator

Store result

Start

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Input Data: 2501- 67h (0110 0111), 2502- 22h (0010 0010)

Output Data: 2503- 45h (0100 0101) Precautions:-


(2) Make sure that all the machine codes/ mnemonics are as per the program. Result:- Subtraction of 22h from 67h is 45h. Hence experiment for the subtraction of two 8-bit BCD numbers has been successfully performed.

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Write a program using 8085 for multiplication of two 8-bit numbers by repeated addition method. Apparatus:-

8085-micro-processor kit (STUDENT-85AD), ASCII (PS-2) keyboard. Theory:

Repeated Addition method of multiplication is simplest method of multiplying two numbers.

For use in microprocessors, we first store multiplicand and multiplier in any of its general purpose

registers. We use a register pair for the storage of result. Then we keep on adding the multiplicand to

result and decrement the multiplier every time we add. This cycle continues until multiplier becomes

zero. Then the result is moved to any desired memory location. A well documented program for this

purpose is given below. Program:-

Memory Machine Code Labels Mnemonics Operands Comments

Address

2000 0E,10 MVI C,08h Move the multiplicand in reg. C

2002 1E,20 MVI E,07h Move the multiplier in reg. E

2004 06,00 MVI B,00 Load 00h in B reg.

2006 21,00,00 LXI H,0000 Initial Product=0000

2009 09 UP1: DAD B HL+BC=>HL

200A 1D DCR E Decrement reg. E

200B C2,09,20 JNZ UP1 Jump if not zero to location up1

200E 22,01,25 SHLD 2501h Store result at 2501h

2012 76 HLT Stop

Procedure:-






[M]<2000>[ENTER].

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press [ENTER].



Flow-Chart of the process:-

Start

Get multiplicand in register C

Get multiplier in register E

Initialize register B. Initialize H-L register-pair for

result storage

Add Multiplicand to H-L register

Decrement multiplier in E register

Is E=0?

Yes Store result

Stop

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Input Data: C reg.- 08h (0000 1000), E reg.- 07h (0000 0111) Output Data: 2501- 38h (0011 1000)

Precautions:-



Result:-

Multiplication of 08h and 07h is 38h. Hence experiment for the multiplication of two 8-bit numbers by repeated addition method has been successfully performed.

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EXPERIMENT NO.9 Aim:-

Write a program using 8085 for multiplication of two 8-bit numbers by bit-rotation method. Apparatus:-

8085- micro-processor kit (STUDENT-85AD), ASCII (PS-2) keyboard. Theory:-

In binary multiplication we see that when a multiplicand is multiplied by 1 the product is

same as the multiplicand. When a multiplicand is multiplied by zero, the product is zero. The

procedure for multiplication is that multiplicand is rotated to left by one bit if the MSB of multiplier

is 1. Partial product is stored. We keep on rotating the multiplier, decrementing cycle-counter and add

the multiplicand to the partial product until the cycle-counter which is set to 8 becomes zero. When

cycle-counter is 0 that means we have completed the multiplication and we store the result as our

final result. The method of multiplication applied here is known as BIT ROTATION METHOD. Program:-


Address Code

2000 2A,01,25 LHLD 7501 H Get Multiplicand in H-L pair.

2003 EB XCHG Exchange HL pair with DE pair

2004 3A,03,25 LDA 7503 H Get 2nd no. in acc.

2007 21,00,00 LXI H,0000 Initial product in HL=00

200A 0E,08 MVI C,08H Count=08 in reg. C

200C 29 LOOP DAD H Shift partial product left

by 1 bit

200D 17 RAL Rotate multiplier by 1bit.

200E D2,12,20 JNC AHEAD If no carry, go to AHEAD

2011 19 DAD D Product=Product +Multiplicand

2012 0D AHEAD DCR C Decrement Count

2013 C2,0C,20 JNZ LOOP If C≠0, go to LOOP

2016 22,04,25 SHLD 7504 Store result

2019 76 HLT Stop

Procedure:-

(1) Connect the kit to the power supply. (2) Connect the PS-2 keyboard to the PS-2 (Female) connector CN-7 of the kit. (3) Switch ON the kit.

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(4) LCD displays “STUDENT-85”, if not so check power connections.

(5) If backlight is glowing and display shows something else press the RESET key provided on the kit. If still the persists contact Lab staff.

(6) To feed the program and data in the RAM, for machine language follow the steps 7 to 9 and

for assembly language programming steps 10 to 14.

(7) Press [M]<Starting Address >[ENTER]. Starting address here is 2000H, so syntax will be [M]<2000>[ENTER].







press [ENTER].



Input Data: C reg.- 08h (0000 1000), E reg.- 07h (0000 0111) Output Data: 2501- 38h (0011 1000) Precautions:-



31

Flow-Chart of Process:-

Start

Get Multiplicand, Get Multiplier

Initial value of Product = 00 Initial Count =08

Shift Product left one Bit Shift Multiplier left one Bit

Is there a No carry

from multiplier

Yes Product = Product + Multiplicand

Count = Count -1

Is

No Count =0?

yes

Store Product

Stop

Result:-


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Write a program using 8085 for division of a number by another number (8-bit) using repeated subtraction method.

Apparatus:-

8085-micro-processor kit (STUDENT-85AD), ASCII (PS-2) keyboard. Theory:-

Repeated subtraction method of division is the simplest method of dividing two numbers. For

use in microprocessors, we can store dividend and divisor in any of its general purpose registers. If

the dividend is bigger than 8-bit we can use a register pair for its storage. Then we keep on

subtracting the divisor form dividend or remainder of subtraction and increment the quotient every

time we subtract. This cycle continues until remainder becomes zero or less than divisor. Then the

result is moved to any desired memory location. A well documented program for this purpose is given

below. Program:-

Memory Machine

Code Labels Mnemonics Operands Comments

Address

2000 21,00,21 LXI H,2100h Address of divisor in H-L pair

2003 46 MOV B,M Divisor in reg. B

2004 23 INX H Increment H-L pair

2005 7E MOV A,M Dividend in Acc.


2007 0E,00 MVI C,00h Initialize C reg. for quotient

storage

2009 B8 LP1 CMP B Compare acc. With B reg.

200A DA,13,20 JC LOOP If carry jump to LOOP

200D 90 SUB B Subtract divisor from dividend

200E 0C INR C Increment C reg.

200F C3,09,20 JMP LP1 Jump back and repeat from to

LP1

2012 77 LOOP MOV M,A Store remainder at 2102h


2014 71 MOV M,C Store result at 2103h

2015 76 HLT Stop

33

Procedure:-






[M]<2000>[ENTER].







press [ENTER].



Input Data: 2100h - 38h (0011 1000), 2101h- 07h (0000 0111) Output Data: 2102h - 00h (0000 0000), 2103h- 08h (0000 1000)

Precautions:-



34


Start

Get divisor in B reg.

Get dividend in accumulator

Initialize C reg. for quotient Compare dividend & divisor

Yes Is B > A?

No Subtract B from Acc.

Increment C reg. and

Go back to comparison

Store remainder from Acc. to memory and quotient from C to memory.

Stop Result:-


35

EXPERIMENT NO.11 Aim:

Write a well document program using 8085 for division of two 8-bit numbers by bit-rotation method. Apparatus:

8085-micro-processor kit (STUDENT-85AD), ASCII (PS-2) Keyboard. Theory:

The computer performs division by trail subtraction. The divisor is subtracted from 8 MSBs

of dividend. If there is borrow, the bit of quotient is set to 1; otherwise 0.to line up the dividend and

quotient properly the dividend is shifted left by one bit before each trail of subtraction. The dividend

and quotient share a 16-bit register. Due to shifting of dividend one bit register falls vacant in each

step. The quotient is stored in vacant bit positions. The program and flow-chart are given below. Program:


Address Code

2000 2A, 01,25 LHLD 2501 H Enter the 16 bit address in HL pair

2003 3A, 03,25 LDA 2503 H Get divisor from 2503

2006 47 MOV B, A Divisor in register B

2007 0E, 08 MVI C, 08 Count = 08 in register C.

2009 29 LOOP DAD H Shift dividend and quotient left by one

bit.

200A 7C MOV A, H Most significant bits of dividend in acc.

200B 90 SUB B Subtract divisor from MSB of dividend.

200C DA,11,24

JC AHEAD Is MSB of dividend>divisor? No, go to

AHEAD.

200F 67 MOV H, A MSB of dividend in reg. H

2010 2C INR L Yes, add 1 to quotient.

2011 0D AHEAD DCR C Decrement count.

2012 C2,09,24 JNZ LOOP Is count=0? No, jump to loop.

2015 22,04,25

SHLD 2504 H Store quotient in 2504 and

remainder in 2505 H.

2018 76 HLT Stop.

36

Procedure:-






[M]<2000>[ENTER].







press [ENTER].



Input Data: 2100h - 38h (0011 1000), 2101h- 07h (0000 0111) Output Data: 2102h - 00h (0000 0000), 2103h- 08h (0000 1000) Precautions:-

(1) Make sure proper handling of equipments/ kits. (2) Make sure that all the machine codes/ mnemonics are as per the program.

37


Start

Get dividend, get divisor Count =08, Quotient =00

Shift dividend left one bit Shift divisor left by one bit

No Is MSBs of

Dividend >

divisor ?

Quotient = Quotient+1 .

8MSBs of dividend =8MSBs of dividend -divisor

Count= count -1

No

Is count =0 ?

Yes

Store remainder and quotient

Stop Result:-

Division of 38h by 07h is 08h. Hence experiment for the division of an 8-bit numbers by repeated addition method has been successfully performed.

38

EXPERIMENT NO.12 Aim:

To write an ALP for finding the square of a number from look up table method using 8085. Apparatus:

8085-micro-processor kit (STUDENT-85AD), ASCII (PS-2) keyboard. Theory:

The squares of number are stored in certain memory locations in tabular form. This table is

called look-up table. Here squares of numbers from 00 to 09 are stored at locations 2600 to 2609h.

All these values are in decimal numbers. The program given below is self explanatory. Program:

Memory Machine Mnemonics Operands Comments

2000 3A,00,25 LDA 2500 H Get number in accumulator

2003 6F MOV L,A Move From A into reg. L

2004 26,26 MVI H,26H Get 26 in reg. H

2006 7E MOV A,M Square of data in accumulator

2007 32,01,25 STA 2501 H Store square in 2501 H.

200A 76 HLT Stop

Look-Up table

Memory Address Data

2600h 00

2601h 01

2602h 04

2603h 09

2604h 16

2605h 25

2606h 36

2607h 49

2608h 64

2609h 81

39

Procedure:-

(1) Connect the kit to the power supply. (2) Connect the PS-2 keyboard to the PS-2 (Female) connector CN-7 of the kit.

(3) Switch ON the kit. LCD displays “STUDENT-85”, if not so check power connections. If

backlight is glowing and display shows something else press the RESET key provided on the

kit. If still the persists contact Lab staff.



[M]<2000>[ENTER].







press [ENTER].



Flow-Chart of the process:- Start

Get number from memory

Form the address of Square from the

Look-up table

Find the square of the number

Store result

Stop

40

Input Data: 2500h- 07d

Output Data: 2503h- 49d Precautions:-



Square of the number 07d is 49d. Hence experiment for the calculation of square of given number using look-up table has been successfully performed.

41

EXPERIMENT NO. 13 Aim:

Write a program using 8085 for finding largest number in a data array. Apparatus:

8085 microprocessor kit (STUDENT-85AD),ASCII (PS-2) Keyboard.

Program:

Memory Machine Mnemonics Comments

Address Code

7000 21,00,75 LXI H, 7500H Address for count in H-L pair

7003 4E MOV C, M Count in register C

7004 23 INX H Address of Ist number in HL Pair

7005 7E MOV A,M Ist no. in accumulator

7006 0D DCR C Decrement count

7007 23 Loop INX H Address of next number

7008 BE CMP M Is next number>previous no.

7009 D2,0D,70 JNC Ahead If not carry,jump to ahead

700C 7E MOV A,M Get larger no. into acc.

700D 0D Ahead DCR C Decrement count

700E C2,07,70 JNZ Loop Jump if no zero to loop.

7011 32,50,74 STA 7450 H Store result at 7450.

7014 CF RST 1 Terminate.

Procedure:-






[M]<2000>[ENTER]. (8) The LCD display will show the address and its content. Modify the content according to the


42





press [ENTER].

Flow Chart:-

43

Input Data 7500-03 (Counter)

7501-

7502-

7503-

Output Data: 7450- Precautions:-

(1)Make sure proper handling of equipments/ kits.


Result:-

Thus the largest number in array stored in respective memory location and verified the data.

44

EXPERIMENT NO.14 AIM:

To study of 8086 Microprocessor Kit. APPARATUS:

VMC-8609 8086 microprocessor trainer kit. THEORY:

The 8086 is a 16-bit, N-channel, HMOS microprocessor. The term HMOS is used for“high-

speed MOS”. The 8086 uses 20 address lines and 16 data lines. It can directly address up to220

=

1Mbytes of memory. The 16-bit data word is divided into a low-order byte and a high-order byte. The 20 address lines are time multiplexed lines. The 16 low-order address lines are time multiplexed with data, and the 4 high-order address lines are time multiplexed with status signals. OPERATING MODES OF 8086

There are two modes of operation for Intel 8086, namely the minimum mode and the

maximum mode. When only one 8086 CPU is to be used in a microcomputer system the 8086 is used

in the minimum mode of operation. In this mode the CPU issues the control signals required by

memory and I/O devices. In case of maximum mode of operation control signals are issued by Intel

8288 bus controller which is used with 8086 for this very purpose. When MN/MX is high the CPU

operates in the minimum mode. When it is low the CPU operates in the maximum mode. Pin Description for Minimum Mode

For the minimum mode of operation the pin MN/MX is connected to 5V d.c.supply. The description of the pins from 24 to 31 for the minimum mode is as follows:

INTA(Output): Pin no. 24 Interrupt acknowledge. On receiving interrupt signal the processor issues an interrupt acknowledge signal. It is active LOW.

ALE(Output) : Pin no. 25 Address latch enable. It goes HIGH during T1. The microprocessor sends this signal to latch the address into the Intel 8282/8283 latch.

DEN(Output) : Pin no. 26 Data enable. When Intel 8286/8287 octal bus transceiver is used this

signal acts as an output enable signal. It is active LOW.

DT/R(Output) : Pin no. 27 Data Transmit/Receive. When Intel 8286/8287 octal bus transceiver is

used this signal controls the direction of data flow through the transceiver. When it is High data are

sent out. When it is LOW data are received.

M/IO(Output) : Pin no. 28.Memory or I/O access. When it is HIGH the CPU wants to access

memory. When it is LOW the CPU wants to access I/O device.

WR (Output): Pin no. 29. Write. When it is LOW the CPU performs memory or I/O write

Operation.

HLDA (Output) : Pin no. 30.HOLD acknowledge. It is issued by the processor when it receives HOLD signal. It is active HIGH signal. When HOLD request is removed HLDA goes LOW.

45

HOLD (Output) : Pin no. 31.Hold. When another device in microcomputer system wants to use

the address and data bus, it sends a HOLD request to CPU through this pin. It is an active HIGH

signal. Pin Description for Maximum Mode

For the maximum mode of operation the pin MN/MX is made LOW. It is grounded. The description of the pins from 24 to 31 is as follows: QS1,QS0(Output): Pin no. 24,25 Instruction Queue status. Logic are given below:

QS1QS0

0 0 No operation

0 1 1st

byte of opcode from queue

1 0 Empty the queue

1 1 Subsequent byte from queue S0,S1,S2(Output) : Pin nos. 26,27,28.status signals. These signals are connected to the bus controller Intel 8288.The bus controller generates memory and I/O access control signals. Table for status signals is:

S2 S1 S0

0 0 0 Interrupt acknowledge

0 0 1 Read data from I/O port

0 1 0 Write data into I/O port

0 1 1 Halt

1 0 0 Opcode fetch

1 0 1 Memory read

1 1 0 Memory write

1 1 1 Passive state.

LOCK(Output) : Pin no. 29.It is an active LOW signal. When it is LOW all interrupts are masked

and no HOLD request is granted. In a multiprocessor system all other processors are informed by this

signal that they should not ask the CPU for relinquishing the bus control.

(Bidirectional) : Pin no. 30,31. Local bus Priority control. Other

RQ / GT1, RQ / GT0

processors ask the CPU through these lines to release the local bus. RQ / GT1 has higher priority

than RQ / GT0

FUNCTIONAL UNITS OF 8086 :

The 8086 contains two functional units: a bus interface unit (BIU) and an execution

unit(EU). The general purpose registers, stack pointer, base pointer and index registers, ALU, flag register(FLAGS), instruction decoder and timing and control unit constitute execution unit(EU). The segment registers, instruction pointer and 6-byte instruction queue are associated with the bus interface unit(BIU).

46

BLOCK DIAGRAM OF 8086:

REGISTERS OF 8086 : The Intel 8086 contains the following registers: a) General Purpose Register

b) Pointer and Index Registers

c) Segment Registers d) Instruction Registers e) Status Flags

Introduction to VMC-8609 (8086 Microprocessor Trainer Kit) General Description:

VMC-8609 is a single board Microprocessor Training/ Development kit. Configured around

the INTEL-8086 microprocessor. This kit can be used to train engineers, to control any industrial

process and to develop software for 8086 system.

The kit has been designed to operate in the Maximum or Minimum mode. Co-processors 8087 or 8089 can be added if required. The 8086 can also be replaced by 8088 microprocessor.

The kit communicates with the outside world through an IBM PC compatible keyboard and LCD display.

47

VMC-8609 is packed up with powerful monitor in 32KB of factory programmed EPROMs

and 32KB of RAM for user. This memory can be expanded up to 256KB each. The system has 72

programmable I/O lines. The serial I/O communication is made possible through 8251.

Component Side Layout of VMC-8609 trainer kit

For control applications three 16-bit timer/counters are available through 8253. For real time

applications, 8 level of interrupt are provided through 8259. VMC-8609 provides onboard battery

backup for RAM. This saves the user program in case of power failure.

The onboard resident SYSTEM MONITOR SOFTWARE is very powerful. It provides

various software commands like BLOCK MOVE, INSERT, DELETE, FILL etc. which are helpful in

debugging/ developing software. An onboard Assembler/ Disassembler is also provided on VMC-

48

8609. The kit also supports MASM.VMC-8609 also has onboard buzzer for self testing of hardware

and software. This kit is also provided with a Centronix Printer port to take out the prints of the

program written in the RAM of kit. A Real-Time Clock is provided onboard for real time

applications. OPERATION:

The operation of the kit is very simple. The operation instructions will be displayed when the device is being switched ON or RESET. After power on the system, it will display as follows:

VMC-8609

ENTER RETURN KEY . . . After pressing [ENTER], the operating commands will be displayed:

VMC - 8 6 0 9 . . 7F F F

> A. D. F. G. I. M. P. T. U.

ASSEMBLE UNASSEMBLE

DUMP TRACE

FILL

PRINT

MOVE

GO

INTERRUPT COMMAND DESCRIPTION: 1) [A] ASSEMBLE : Press the key [A] and the LCD is shown as:

VMC - 8 6 0 9 . . 7F F F A

Now the user enters the segment address and effective address simultaneously as follows:

VMC - 8 6 0 9 . . 7F F F

A 0 0 0 0 : 0 4 0 0

Now press [ENTER] key, the effective address will appear

0 4 0 0

From now onwards user can type the program in assembly language as shown below:

49

0 4 0 0 M O V B L , C 0

Pressing [ENTER] from here will store the instruction at the effective address specified in

the instruction and display will show next address. Then next instruction can be typed and stored in

same manner. The sequence will continue till the end of the particular program. 2) [D] DISPLAY OR MODIFY THE RAM’S CONTENT:

Press [RESET] and [ENTER] to return to the command menu. Select [D] from this menu the display will show as follows

VMC - 8 6 0 9 . . 7F F F

D

Type the segment address and offset address of the desired location you wish to modify

VMC - 8 6 0 9 . . 7F F F D 1 2 3 4 : 0 3 0 0

(NOTE: Segment address will be taken as 0000H until and unless specified.) Press [ENTER], the display will show the content of RAM as shown below:

Segment Offset Content of total

Base Address 8 bytes of RAM

1 2 3 4 : 0 3 0 0 B 3 C 0 9 A 7 8 F 0 0 0 F 0 B B

From here on content of any RAM location can be modified. UP/ DOWN and RIGHT/ LEFT arrow keys can be used to go to next or previous memory location. 3) [F] FILL CONSTANT DATA INTO RAM:

Press [RESET] and [ENTER] to return to the command menu. Select [F] from this menu the display will show as follows:

VMC - 8 6 0 9 . . 7F F F

F Type the address and data in the manner shown below:

VMC - 8 6 0 9 . . 7F F F F 0 0 0 0 : 0 4 0 0 0 4 5 0 5 7

50

Segment Starting Ending Constant

Base Address Address to be filled

Press [ENTER] and the value 57H will be stored at offset 0400H to 0450H. 4) [G] GO FOR RUN/ EXECUTION:

Press [RESET] and [ENTER] to return to the command menu. Select [G] from this menu the display will show as follows:

VMC - 8 6 0 9 . . 7F F F G

Type the Segment Address and Offset Address of the starting point of the program.

Press [ENTER] [F7] and [ENTER] again thereafter in order to RUN the program. The code will run

and the end results will be stored at the memory locations specified in the program which can be

verified using [D] command. 5) [I] INTERRUPT:

Press [RESET] and [ENTER] to return to the command menu. Select [I] from this menu the display will show as follows:

VMC - 8 6 0 9 . . 7F F F I N T P : 0 0 0 0 . 0 0 0 0 . 0 0 0 0

First Int. Second Third Int.

Offset Int. Offset Offset

Address Address Address

Three INTERRUPTs (Offset Addresses) can be set in for the program execution, the CPU

will continuously make a single step subprogram for checking IP values. When the IP registers has

the same value as the INTERRUPTs addresses, it will enter the INTERRUPT’s subprogram.

6) [M] MOVING DATA:

The command MOVE is used to move data in the memory from a specified address to another

address by inputting the starting address, the ending address and the desired target address. A return

key is then used to execute the changes.

51

Press [RESET] and [ENTER] to return to the command menu. Select [M] from the menu, the display will show as follows:

Segment Starting Ending

Source Base Offset Offset

M S – 0 0 0 0 : 0 0 0 0 8 0 8 8

T – 0 0 0 0 : 0 0 0 0

Target

Segment Starting

Base Offset

7) [P] PRINT:

This command allows the user to print the output. Printer to be connected to the I/O port of 8255.

Press [RESET] and [ENTER] to return to the command menu. Select [P] from the menu, the display will show as follows:

PRESS S / F FOR SPEED

If the printer is not connected at this time, press key [P] so the LCD would show ‘PRINTER

ERROR!’ in the second line. Here S/F indicates ‘SLOW’ or ‘FAST’. Now according to the printer,

press ‘S’ or ‘F’ key, then following display will be shown on the screen.

PRINTER ON ! >B . D . U .

UNASSEMBLE

DUMP

BUFFER

Description of the above commands is as follows:

[D]…..command will allow the printer to print the machine code.

[U]…..command will allow the printer to print the assembly program.

[B]…..command sends data of RAM directly to the printer without going through any modification.

52

The starting address and ending address need to be entered first before using commands [B], [D] and [U]. Followed by an enter or ARROW UP key in order to print the output. 8) [T] TRACE PROGRAM (AN N-STEP DESIGNED COMMAND):

This command is used for program execution in any desired number of steps. TRACE

will enter the interrupt subprogram every time the program is executed. N has a decimal range from

1-99 with 10 as the rounding off number and only operates if N is not 0; otherwise it will clear the

function.

After feeding the program, Press [F7] key, then menu display will show as follows:

0 4 0 E

>A. D. F. G. I. M. P. T. U.

Press the key [T], the screen displays as follows:

0 4 0 E T 0 0 - STEP

Here user can enter any number of steps from 00 to 99 and start execution using [G]

command. For example, 01 is entered in place of 00 in the above command the program will be

interrupted after execution of every instruction and if 02 entered in place of 00 then the program will

be interrupted after execution of every two instructions and so on. 9) [U] UNASSEMBLE:

The UNASSEMBLE command decodes the value of a group of memory location mnemonics and displays on the display.

Press [RESET] and [ENTER] to return to the command menu. Select [U] from the menu, the display will show as follows:

VMC - 8 6 0 9 . . 7F F F U 0 4 0 0

Here 0400 is the address from where the assembly language code will be unassembled. Press [ENTER] and the display will show as:

Effective Address Machine Code

0 0 0 0 : 0 4 0 0 B 0 3 0

MOV A L , 3 0

Assembly mnemonics

53

User can use UP/ DOWN arrow keys to go to previous or next address respectively. MEMORY MAPPING:

Memory Locations Usage

0000:0000 – 0000:FFFF RAM AREA (Odd and Even RAM)

F000:0000 – F000:FFFF ROM AREA (Odd and Even ROM)

RAM MEMORY

Memory Locations Usage

Interrupt Vector Section (INT1, INT2,

INT3) have the arranged Interrupt Section

0000:0000 and Stack Segment.

0000:0390 Buffer

0000:039B System Data

0000:93E0 Buffer (Only if needed)

0000:0400 – 0000:7FFF User RAM Area

I/O MAPPING

Device Name Port Name Port Address Connector

Port-A 70

Port-B 72

8255-I

CN3

Port-C 74

CWR 76

Port-A 80

Port-B 82

8255-II

CN4

Port-C 84

CWR 86

Port-A 10

54

Port-B 12

8255-III

CN5

Port-C 14

CWR 16

Counter 0 00

Counter 1 02

8253

CN6

Counter 2 04

Counter 3 06

Keyboard Input 20

Latch

Output 22

Data Word 30

8259

CN6

Command Word Register 32

Data Word Register 50

8251

CN7

Command Word Register 52

RESULT:

Basic architecture of 8086 and operation of VMC-8609 Microprocessor trainer kit successfully studied.

55

EXPERIMENTNO. 15 AIM:

Write a program using 8086 for addition of two 16 bit numbers. APPARATUS:

8086 microprocessor kit, 5V power supply, Keyboard. Program:

Memory Machine Mnemonics Operands Comments

Address Code

1000 B8,34,12 MOV AX,1234 Load 1234 in AX

1003 BA,65,87 MOV DX,8765 Load 8765 in DX

1006 03,C2 ADD AX,DX Add DX with AX

1008 8B,C8 MOV CX,AX Move answer to CX

1009 CD,A5 INT A5 Jump to command mode saving all registers.

CIRCUIT DIAGRAM / BLOCK DIAGRAM:-

START

Get the first No.

Get the second No.

Add. Two numbers

Store the result

END PROCEDURE:-

56

1) Switch on the kit.

2) Go to assembler then type the mnemonics of our program in respective memory location.

3) Ensure the input weather placed in respective memory location.

4) Execute the program with starting address.

5) Verify the output data in respective memory location. INPUT DATA

1000-1234(H)

1001-8765(H) OUTPUT DATA

AX -9999(H) PRECAUTIONS:-

Make sure that all the machine codes should be as per specified in the program. RESULT:-

Thus the addition of two 16 bit numbers has performed and verified.

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